For the illustrated circuit with 4 μF capacitor, the steady-state charge is q₀ = 16 μC. The battery is then removed and nodes A and C are shorted. If the time for charge to fall to 4 μC is X μs and the initial battery emf is Y V, calculate XY/(96 ln 2).
Physics · Mixed Chapter
JEE Mains

Question

For the illustrated circuit with 4 μF capacitor, the steady-state charge is q₀ = 16 μC. The battery is then removed and nodes A and C are shorted. If the time for charge to fall to 4 μC is X μs and the initial battery emf is Y V, calculate XY/(96 ln 2).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Explanation: Initially the capacitor voltage is 16 μC/4 μF=4 V. The resistor dividers create V at B−V at D=E/6, so E=24 V (Y=24). After battery removal and A–C shorting, R along B–A=(2+1)|| (3+3)=2 Ω, while R along D–A=4||4=2 Ω; total Rₑᵩ=4 Ω. Thus τ=Rₑᵩ C=16 μs. Falling from 16 to 4 μC takes X=τ ln4=32 ln2 μs. XY/(96 ln2)=(32 ln2×24)/(96 ln2)=8.